How to prove this inequality in interval: (0,π2) 2cos⁡x+sec2x−3≥0

Restrict the domain to some meaningful interval such as ${\displaystyle x\in (-\frac{\pi }{2},\frac{\pi }{2})}$ so that ${\displaystyle \cos x>0}$. In that case, using the AM-GM inequality you have:
${\displaystyle LHS=\cos x+\cos x+{\sec }^{2}x-3\ge 3\sqrt[3]{\cos x\cdot \cos x\cdot {\sec }^{2}x}-3=3-3=0=RHS}$

We have
${\displaystyle 2\cos x+{\sec }^{2}x-3=2(\cos x-1)+(\frac{1}{{\cos }^{2}x}-1)=\frac{(2\cos x+1){(\cos x-1)}^2}{{\cos }^{2}x}}$
Thefore the inequality does not hold.

By Fermat's theorem, we examine zeroes of derivative of the function $f(x)=2\cos x+{\sec }^{2}x-3$ to find the extrema, that is, to solve the following:
$f\prime (x)=2\tan x{\sec }^{2}x-2\sin x=0$
The solution is $x=\{k\pi |k\in \mathbb{Z}\}$. When $x=\pi$, we get
$2\cos (x)+{\sec }^2(x)-3=-4$
Thus $2\cos x+{\sec }^{2}x-3\ge 0$ doesn't hold.
Since the interval is added, we can prove the claim is true. For x=0, f(x)=0. And when $x\in (0,\frac{\pi }{2})$, it is trivial that $f\prime (x)\ge 0$, thus f(x) monotonously increases on $(0,\frac{\pi }{2})$. Hence $f(x)\ge f(0)=0\text{ for all }x\in (0,\frac{\pi }{2})$ and we finished the proof.